[meteorite-list] meteoroid temperature- Assumptions (P. II)

From: MexicoDoug_at_aol.com <MexicoDoug_at_meteoritecentral.com>
Date: Fri Jan 7 05:32:44 2005
Message-ID: <ae.6931b2af.2f0fbf47_at_aol.com>

Part II

Assumptions & Discussion:
T = [(absorptivity/emissivity)*(Energy flux/sigma)*(a/A)]^(1/4)
where:
Emissivity = energy ratio emitted at a temperature = compositional property.
Absorptivity = energy fraction absorbed at a temperature = compositional
property.

Temperature is proportional to absorptivity but proportional to the inverse
of emissivity, i.e. T^(4)=k*(absorptivity/emissivity).

The useful form of this law is called the Stephan-Boltzmann Law: Energy =
sigma*T^4 (S-B law constant, sigma = 5.67*10^-8*Wm^-2*K^-4)
That the integrated energy radiated from a black booty is proportional to
the fourth power of the temperature. This "law" is useful to convert the
energy hitting an object into the temperature and is really the scientific key to
address your question approximately.

Kirchoff's (other) Law:
For a body in radiative equilibrium energy absorption = energy emission. So
we consider the very plausible scenario that the meteoroid's position in
orbit doesn't alter radically (it doesn't travel near light speed!)
Solar energy is mainly provided for absorption in the UV-Visible range.
Energy is emitted in the IR range (vibrational energy, the meteoroid doesn't
emit much light energy:-).
Meteoroide is spherical in shape (OK not generality, so could vary maybe 50%
either way for example when a planar shaped meteoroid had verrry low
rotational energy w/r to the Sun)

ABSORPTIVITY = CC = 0.8, OC = 0.65, Fe-Ni = 0.80, "ideal" = 1
EMISSIVITY = CC = 0.88, OC = 0.85, weathered Fe-Ni = 0.28, "ideal" = 1

Assumed constant emissivities, but actually they vary, for example,
decreasing somewhat (and then only by a square root) as the AU's increase for Iron,
and a lot for Nickel, though I expect the "weathered" surface might somewhat
mitigate this.

The temperature of a meteoroid in space will of course depend on the
latitude, depth, it's cross sectional exposure vs. overall mass distribution and
distance to the Sun, just like any other non-radioactive cooled body whether in
orbit or passing through.
 
Proximamente: More Assumptions & Discussion
 
Received on Fri 07 Jan 2005 05:32:39 AM PST